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\begin{document}

\begin{frontmatter}

\title{On the explicit structure of $K_2$ of finite abelian group ring over finite field}


%% Group authors per affiliation:
\author[address]{Guoping Tang\corref{correspondingauthor}}
\ead{tanggp@ucas.ac.cn}

\author[address]{Hang Liu}
\ead{h2.liu@vu.nl}


\address[address]{School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China}

\cortext[correspondingauthor]{Corresponding author}

\begin{abstract}
Let $\mathbb{F}$ be a finite field
and $G$ be a finite abelian group. We give
the explicit structure of $K_2(\mathbb{F}G)$.
\end{abstract}

\begin{keyword}
$K_2$-group, finite abelian group
ring.
\end{keyword}

\end{frontmatter}




\section{Introduction}
Let $p$ be a prime number and $\mathbb{F}$ be the finite field of order $p^f$.
By (3.1) in~\cite{GT} we have the following decomposition
formula:
\begin{equation}\label{eqn:decompsition}
K_2(\mathbb{F}[G\times C_{p^m}])\cong K_2(\mathbb{F}G)\oplus
K_2(\mathbb{F}G[t]/(t^{p^m}),(t)).
\end{equation}
where $G$ is a finite abelian group and $C_{p^m}$ a cyclic group of
order $p^m$. By Theorem 15 of~\cite{K}, $K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ can be expressed in terms of Dennis-Stein symbols and relations.
More generally, let $R$ be a commutative ring with unit and $I\subseteq rad(R)$.
The relative $K_2$-group $K_2(R,I)$ is generated by
Dennis-Stein symbols $\langle a,b\rangle $ with $a$ or $b$ in $I$,
satisfying the following relations:
\[\begin{array}{lll}
(DS1)\quad\;\langle a,b\rangle=-\langle b,a\rangle & \qquad \textrm{if }\; a\in I;\\
(DS2)\quad\;\langle a,b\rangle+\langle a,c\rangle=\langle
a,b+c-abc\rangle & \qquad\textrm{if }\; a\in I\; \textrm{or}\; b,c \in I;\\
(DS3)\quad\;\langle a,bc\rangle=\langle ab,c\rangle+\langle
ac,b\rangle & \qquad\textrm{if }\; a\in I.\end{array}\]
By Lemma 1.5 of~\cite{S} these relations imply
\begin{equation}\label{eqn:mrelation}
\langle a,b^m\rangle =m\langle ab^{m-1},b\rangle.
\end{equation}
Moreover if $R$ is an $\mathbb{F}_p$-algebra, these relations also imply
\begin{equation}\label{eqn:prelation}
p^r\langle a,b\rangle=\langle a^{p^r}b^{p^r-1},b\rangle.
\end{equation}
Using these symbols and relations, Chen, Gao and Tang~\cite{CGT} found a small generating set of
$K_2(\mathbb{F}_pG[t]/(t^{p^m}),(t))$.
Moreover, the order of
$K_2$ of finite abelian group ring over finite field was given by Proposition~6.3 of~\cite{O}, so the order of
$K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ can be easily calculated by~\eqref{eqn:decompsition}.
Combining the information of the order of $K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$
and the order of the group generated by the generating set, they found the structure
of $K_2(\mathbb{F}_p G[t]/(t^{p^m}),(t))$ and subsequently the structure of $K_2(\mathbb{F}_pG)$.

But for an arbitrary finite field $\mathbb{F}$,
we only know the order of $K_2(\mathbb{F}G)$.
In this paper we will give the explicit structure of the group and subsequently
the structure of $K_2$ of finite abelian group ring over finite field.
More precisely, we prove the following theorem by modifying the proof in~\cite{CGT}.
\begin{thm}\label{thm:main}
Let $\mathbb{F}$ be a finite field of characteristic $p$ with $p^f$ elements. Let $G=G_p \oplus H$
be a finite abelian group where $G_p$ is the Sylow $p$-subgroup of $G$ which has exponent $p^e$.  Let $r_i$ denote the $p^i$-rank of $G_p$.
Then
$$K_2(\mathbb{F}G)=C_{p^e}^{f(r_e-1)(|G^{p^{e-1}}|-|G^{p^{e}}|)}\bigoplus_{i=1}^{e-1}C_{p^i}^{f(r_i-1)(|G^{p^{i-1}}|-|G^{p^i}|)-(r_{i+1}-1)(|G^{p^i}|-|G^{p^{i+1}}|)}.$$
\end{thm}







\section{Proof of theorem~\ref{thm:main}}
\begin{lem}\label{lem:vanish}
Let $a$ be an element of $\mathbb{F}G[t]/(t^{p^m})$ and $\alpha$ be an element of $\mathbb{F}$,
then $\langle at^i, \alpha \rangle=0$ in $K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ for $i\geqslant 1$.
Consequently, we have $\langle at^i, \alpha x \rangle = \langle \alpha at^i, x \rangle$ for $x \in \mathbb{F}G[t]/(t^{p^m})$.
\end{lem}
\begin{proof}
If $\alpha=0$, it is obvious $\langle at^i, \alpha \rangle=0$. If $\alpha \neq 0$,
we have $\alpha^{p^m-1}=1$, by~\eqref{eqn:mrelation} and~\eqref{eqn:prelation}
$$\langle at^i, \alpha \rangle = \langle at^i, \alpha^{p^m} \rangle = p^m\langle at^i, \alpha \rangle
=\langle (at^i)^{p^m}\alpha^{p^m-1}, \alpha \rangle = \langle 0, \alpha \rangle = 0.$$
So by (DS3), we have
$$\langle at^i, \alpha x \rangle = \langle xat^i, \alpha  \rangle + \langle \alpha at^i, x \rangle = \langle \alpha at^i, x  \rangle.$$
\end{proof}

The next Theorem~\ref{thm:generator} and Lemma~\ref{lem:main} reduce the number of Dennis-Stein symbols
needed to generate $K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ which are generalizations of Theorem 3.4
and Lemma 3.5 of~\cite{CGT}. The proof is almost identical to corresponding results in~\cite{CGT} except
we use Lemma~\ref{lem:vanish} at certain places. We give the proof here for the sake of completeness.
\begin{thm}\label{thm:generator}
Let $G=\langle\sigma_1\rangle\times \cdots \times \langle
\sigma_n\rangle$ be a finite abelian $p$-group and $\delta_1, \delta_2, \cdots ,\delta_{m}$
be a basis of $\mathbb{F}$ over $\mathbb{F}_p$. Then
$K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ can be generated by
$$S=\{\langle \delta_j gt^k,t\rangle,\langle \delta_j gt^k,\sigma_i\rangle |
g\in G,1\leq i\leq n, 1\leq j\leq m, 1\leq k< p^m\}.$$
\end{thm}
\begin{proof}
By Proposition 1.7 in~\cite{S}, $K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ is
generated by elements $\langle at^i,t\rangle$ and $\langle
at^i,b\rangle$ with $a$, $b\in \mathbb{F}G$ and $1\leq i \leq
p^m-1$, and we define a filtration on
$K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ using these elements. Let
$F_0=0$ and
\begin{itemize}
\item[(1)] when $1\leq k\leq p^m-1$,\\
$F_k=$subgroup generated by $F_{k-1}$ and symbols of type $\langle
at^{p^m-k},t\rangle$;
\item[(2)] when $p^m\leq k\leq 2p^m-2$,\\
$F_k=$subgroup generated by $F_{k-1}$ and symbols of type $\langle
at^{2p^m-k-1},b\rangle$.
\end{itemize}
Then $F_0\subseteq F_1 \subseteq\cdots \subseteq F_{p^m-1}\subseteq
F_{p^m}\subseteq\cdots\subseteq
F_{2p^m-2}=K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$. To prove the
theorem, it suffices to prove that the image of $S\cap F_k$ under
the natural map $F_k\rightarrow F_k/F_{k-1}$ is a set of generators
of $F_k/F_{k-1}$.

By the first part of the proof of Theorem~3.4 in~\cite{CGT},
for $1\leq k\leq p^m-1$, $F_k/F_{k-1}$ is generated
by the image of $\langle \delta_j gt^{p^m-k},t\rangle$ with $1\leq j\leq m$ and $g\in G$.

By the second part of the proof of Theorem~3.4 in~\cite{CGT},
for $p^m\leq k\leq 2p^m-2$, $F_k/F_{k-1}$ can be generated by $\overline{\langle
at^{2p^m-k-1},\delta_j\sigma_1^{h_1}\cdots \sigma_n^{h_n}\rangle}$.
But by Lemma~\ref{lem:vanish} and (DS3), we have
\begin{align*}
\langle at^{2p^m-k-1},\delta_j\sigma_1^{h_1}\cdots
\sigma_n^{h_n}\rangle
=& \langle \delta_jat^{2p^m-k-1},\sigma_1^{h_1}\cdots
\sigma_n^{h_n}\rangle \\
=&
\sum_{i=1}^nh_i\langle
\delta_jat^{2p^m-k-1}\sigma_1^{h_1}\cdots\sigma_i^{h_i-1}\cdots\sigma_n^{h_n},\sigma_i\rangle
.
\end{align*}
So $F_k/F_{k-1}$ is generated by
symbols in $\{\overline{\langle at^{2p^m-k-1},\sigma_i\rangle}|a\in
\mathbb{F}G, 1\leq i\leq n\}$.
By the second part of the proof of Theorem~3.4 in~\cite{CGT} again,
we see
$\overline{\langle at^{2p^m-k-1},\sigma_i\rangle}$ is generated by
symbols $\{\overline{\langle \delta_j gt^{2p^m-k-1},\sigma_i\rangle}|g\in
G\}$.  Now the theorem is proved.
\end{proof}


\begin{lem}\label{lem:main}
Let $\mathbb{F}$, $G$ be as in Theorem~\ref{thm:generator}. Let $\beta,\beta^p,\cdots,\beta^{p^{m-1}}$ be a basis of $\mathbb{F}$ over $\mathbb{F}_p$ and $S$ be as in Theorem~\ref{thm:generator} for this basis. Let
\begin{align*}
T_1 &=\{\langle \beta^{p^j} g^pt^k,t\rangle | g\in G,0 \leq j<m, 1\leq k< p^m
\};\\
T_2 &=\{\langle \beta^{p^j}\sigma_1^{l_1}\cdots
\sigma_i^{l_i}g^pt^k,\sigma_i\rangle | 0\leq l_1,\cdots ,l_{i-1}
\leq p-1, 0\leq l_i\leq p-2, \\& \indent g\in G,1\leq i\leq n, 0 \leq j<m, 2\leq k< p^m\};\\
T_3 &=\{\langle \beta^{p^j}\sigma_i^{p-1}g^pt^k,\sigma_i\rangle | g\in G,1\leq i\leq
n, 0 \leq j<m,2\leq k<p^m, k\equiv 0\mod{p}\}.
\end{align*}
Set $T=T_1\cup T_2 \cup T_3$.  Then
$K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ can be generated by
Dennis-Stein symbols in $S\setminus T$.
\end{lem}
\begin{proof}
By Theorem~\ref{thm:generator},we need only to show that the symbols in $T$ is a sum
of symbols in $S\setminus T$.
Now we consider the symbols in $T_1$,
$T_2$ and $T_3$ respectively.

\textcircled{1} Symbols in $T_1$.\\
Let $g=\sigma_1^{l_1}\cdots \sigma_n^{l_n}$. If $k+1\not \equiv 0
\mod {p}$,
\begin{gather*}
\begin{split}
\langle \beta^{p^j}g^pt^k,t\rangle
&=\langle \beta^{p^j}g^p, t^{k+1}\rangle+\langle t^k,\beta^{p^j}g^pt\rangle\\
&=-\langle \beta^{p^j}t^{k+1},g^p\rangle - \langle \beta^{p^j}g^pt,t^k\rangle\\
&=-p\langle \beta^{p^j}g^{p-1}t^{k+1},g \rangle -k\langle \beta^{p^j}g^pt^k,t\rangle,
\end{split}
\end{gather*}
so we have
\begin{equation*}(1+k)\langle \beta^{p^j}g^pt^k,t\rangle=-p\langle
\beta^{p^j}g^{p-1}t^{k+1},g\rangle=-p \sum_{i=1}^nl_i\langle
\beta^{p^j}g^p\sigma_i^{-1}t^{k+1},\sigma_i\rangle.
\end{equation*}
Since $k+1\not\equiv0\mod {p}$, $\langle
\beta^{p^j}g^p\sigma_i^{-1}t^{k+1},\sigma_i\rangle \in S\setminus T$. By
Proposition 1.8 in \cite{S}, $K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$
is $(k+1)$-divisible, so $\langle \beta^{p^j}g^pt^k,t\rangle$ is a sum of
symbols in $S\setminus T$ when $k+1\not\equiv 0 \mod {p}$.

If $k+1\equiv 0 \mod{p}$, let $k=lp-1$, $1\leq l\leq p^{m-1}$, then
$\langle \beta^{p^j}g^pt^k,t\rangle=\langle \beta^{p^j}g^pt^{lp-1},t\rangle$.
If $l=1$,
\begin{equation*}
\langle \beta^{p^j}g^pt^{p-1},t\rangle=p\langle \beta^{p^{j-1}}g,t\rangle=-p\langle
t,\beta^{p^{j-1}}g\rangle=-p\langle
\beta^{p^{j-1}}t,g\rangle=-p\sum_{i=1}^nl_i\langle \beta^{p^{j-1}}g\sigma_i^{-1}t,
\sigma_i\rangle.
\end{equation*}
If $j=0$, we use $\beta^{p^{-1}}$ to denote $\beta^{p^{m-1}}$ above because $\beta^{p^{m}} = \beta$, 
we will also use this convention below.
Obviously $\langle \beta^{p^{j-1}}g\sigma_i^{-1}t, \sigma_i\rangle \in S\setminus
T$, and $\langle \beta^{p^j}g^pt^{p-1},t\rangle$ is a sum of symbols in
$S\setminus T$.

If $l>1$ and $g\not\in G^p$,
\begin{equation*}
\langle \beta^{p^j}g^pt^{lp-1},t\rangle=p\langle \beta^{p^{j-1}}gt^{l-1},t\rangle.
\end{equation*} and $\langle \beta^{p^{j-1}}gt^{l-1},t\rangle\in S\setminus T$.

If $l>1$ and $g\in G^p$, let $g=g'^p$ for some $g'\in G$, then
\begin{equation}
\langle \beta^{p^{j}}g^pt^{lp-1},t\rangle=p\langle
\beta^{p^{j-1}}g'^pt^{l-1},t\rangle.\label{E3-3}
\end{equation}
Since $\langle \beta^{p^{j-1}}g'^pt^{l-1},t\rangle\in T_1$, we can repeat the
discussions above to show that either $\langle \beta^{p^{j-1}}g'^pt^{l-1},t\rangle$
is a sum of symbols in $S\setminus T$ or $\langle
\beta^{p^{j-1}}g'^pt^{l-1},t\rangle=p\langle \beta^{p^{j-2}}g''^pt^{l'-1},t\rangle$ by
$(\ref{E3-3})$ when $g'\in G^p$ and $l=l'p, l' \geq 1$.  But
$l'-1<l-1<lp-1$, so after a finite number of steps we can show
$\langle \beta^{p^{j}}g^pt^{lp-1},t\rangle$ is a sum of symbols in $S\setminus T$
when $g\in G^p$ and $l>1$.

\textcircled{2} Symbols in $T_2$.\\
 Let $g=\sigma_1^{l_1'}\cdots
\sigma_n^{l_n'}$, then
\begin{gather*}
\begin{split}
\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots \sigma_i^{l_i}g^pt^k,\sigma_i\rangle
&=-\langle \sigma_i,\beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i}g^pt^k\rangle\\
&=-\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i+1}g^p,t^k\rangle-\langle
\sigma_it^k,\beta^{p^{j}}\sigma_1^{l_1}\cdots \sigma_i^{l_i}g^p\rangle\\
&=-\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i+1}g^p,t^k\rangle-\langle
\beta^{p^{j}}\sigma_it^k,\sigma_1^{l_1}\cdots \sigma_i^{l_i}g^p\rangle\\
&=-k\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i+1}g^pt^{k-1},t\rangle-p\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i+1}g^{p-1}t^k,g\rangle\\
&\quad-\langle
\beta^{p^{j}}\sigma_ig^pt^k,\sigma_1^{l_1}\cdots\sigma_i^{l_i}\rangle.
\end{split}
\end{gather*}
For the last two terms in the above equation,
\begin{gather*}
\begin{split}
\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots \sigma_i^{l_i+1}g^{p-1}t^k,g\rangle
&=\sum_{r<i}l_j'\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots \sigma_r^{l_r-1}\cdots
\sigma_i^{l_i+1}g^pt^k,\sigma_r\rangle +l_i'\langle
\beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i}g^pt^k,\sigma_i\rangle\\
&+\sum_{r>i}l_j'\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i+1}\sigma_r^{-1}g^pt^k,\sigma_r\rangle,
\\
\langle\beta^{p^{j}}\sigma_ig^pt^k,\sigma_1^{l_1}\cdots \sigma_i^{l_i}\rangle
&=\sum_{r<i}l_j\langle\beta^{p^{j}}\sigma_1^{l_1}\cdots \sigma_r^{l_r-1}\cdots
\sigma_i^{l_i+1}g^pt^k,\sigma_r\rangle +l_i\langle
\beta^{p^{j}}\sigma_1^{l_1}\cdots \sigma_i^{l_i}g^pt^k,\sigma_i\rangle,
\end{split}
\end{gather*}
so we have
\begin{eqnarray*}
&&(1+pl_i'+l_i)\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i}g^pt^k,\sigma_i\rangle \\
&&=-k\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i+1}g^pt^{k-1},t\rangle-
\sum_{r<i}l_j\langle
\beta^{p^{j}}\sigma_1^{l_1}\cdots \sigma_r^{l_r-1}\cdots
\sigma_i^{l_i+1}g^pt^k,\sigma_r\rangle\\
&&-p\big(\sum_{r<i}l_j'\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_r^{l_r-1}\cdots \sigma_i^{l_i+1}g^pt^k,\sigma_r\rangle
+\sum_{r>i}l_j'\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i+1}\sigma_r^{-1}g^pt^k,\sigma_r\rangle\big).
\end{eqnarray*}
Since $1\leq l_i+1\leq p-1$, and $\sigma_i^{l_i+1}\not\in G^p$,
 $\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i+1}g^pt^{k-1},t\rangle$, $\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_r^{l_r-1}\cdots \sigma_i^{l_i+1}g^pt^k,\sigma_r\rangle(r<i)$
and $\langle\beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i+1}\sigma_r^{-1}g^pt^k,\sigma_r\rangle(r>i)$ in the
above equation are in $S\setminus T$.  And because
$(1+pl_i'+l_i)\not\equiv 0\mod{p}$,
$K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ is $(1+pl_i'+l_i)$-divisible,
so $\langle \beta^{p^{j}}\sigma_1^{l_1}\cdots
\sigma_i^{l_i}g^pt^k,\sigma_i\rangle$ in $T_2$ is a sum of symbols
in $S\setminus T$.

\textcircled{3}Symbols in $T_3$.\\
Let $k=lp$, $1\leq l\leq p^{m-1}-1$, then
$$\langle \beta^{p^{j}}\sigma_i^{p-1}g^pt^{lp},\sigma_i\rangle=p\langle \beta^{p^{j-1}}gt^l,\sigma_i\rangle.$$
If $\langle \beta^{p^{j-1}}gt^l,\sigma_i\rangle\in T_2$, by
the discussions in \textcircled{2}, $\langle \beta^{p^{j-1}}gt^l,\sigma_i\rangle$
is a sum of symbols in $S\setminus T$; If $\langle
\beta^{p^{j-1}}gt^l,\sigma_i\rangle\in T_3$, then $\langle
\beta^{p^{j-1}}gt^l,\sigma_i\rangle=p\langle \beta^{p^{j-2}}g't^{l'},\sigma_i\rangle$ and
$l'<l<lp$. But if $l=1$, $\langle \beta^{p^{j-1}}gt^l,\sigma_i\rangle\in S\setminus
T$. so after a finite number of steps, we can show $\langle
\beta^{p^{j}}\sigma_i^{p-1}g^pt^{lp},\sigma_i\rangle$ is a sum of Dennis-Stein
symbols in $S\setminus T$. Now the lemma is proved.
\end{proof}

Now we have the following theorems which are similar to Theorem 1.1 and Theorem 1.2 of~\cite{CGT}.
\begin{thm}
Let $\mathbb{F}$ be a finite field of order $p^f$ and
$G=C_{p^{\alpha_1}}\times \cdots \times C_{p^{\alpha_n}}=\langle
\sigma_1\rangle\times\cdots\times\langle \sigma_n\rangle$ be a
finite abelian $p$-groups and $\alpha_1,\cdots,\alpha_n\geq m$. Then
$$K_2(\mathbb{F}G[t]/(t^{p^m}),(t))\cong
C_{p^m}^{f|G^{p^m}|(np^{n+1}-(n-1)p^n-1)}\bigoplus_{i=1}^{m-1}C_{p^i}^{f|G^{p^i}|(np^{m-n-i-1}(p^{n+1}-1)^2-(n-1)p^{-n}(p^n-1)^2)}
.$$
\end{thm}
\begin{proof}
Let $\beta,\beta^p,\cdots,\beta^{p^{m-1}}$ be a basis of $\mathbb{F}$ over $\mathbb{F}_p$ as in Lemma~\ref{lem:main}.
Then the proof is the same as Theorem 1.1 of~\cite{CGT} except we use
$\langle \beta^{p^j}\sigma_1^{l_1}\cdots\sigma_n^{l_n}t^k,x\rangle$
instead of
$\langle \sigma_1^{l_1}\cdots\sigma_n^{l_n}t^k,x\rangle$ and
the order of all the sets here is $f$ times of the sets in~\cite{CGT}.
\end{proof}
\begin{thm}\label{thm:mainp}
Let $\mathbb{F}$ be a finite field of order $p^f$ and $G$ a finite
abelian $p$-group of exponent $e$.  Let $r_i$ denote the $p^i$-rank
of $G$.  Then
$$K_2(\mathbb{F}G)=C_{p^e}^{f(r_e-1)(|G^{p^{e-1}}|-1)}\bigoplus_{i=1}^{e-1}C_{p^i}^{f\{(r_i-1)(|G^{p^{i-1}}|-|G^{p^i}|)-(r_{i+1}-1)(|G^{p^i}|-|G^{p^{i+1}}|)\}}.$$
\end{thm}
\begin{proof}
The proof is the same as Theorem 1.2 of~\cite{CGT} except
the number of the cyclic $p$-groups in $K_2(\mathbb{F}G[t]/(t^{p^m}),(t))$ and $K_2(\mathbb{F}G)$ is $f$ times of the number in~\cite{CGT}.
\end{proof}
Now we can prove Theorem~\ref{thm:main} by applying Theorem~\ref{thm:mainp}.

\noindent\textbf{Proof of Theorem~\ref{thm:main}.}
Since $\mathbb{F}H$ is a semisimple ring, by Artin-Wedderburn theorem we have $\mathbb{F}H=\bigoplus_{i=1}^{n}\mathbb{F}_i$ where $\mathbb{F}_i$ is finite field of order $p^{f_i}$.
Since
$$|\mathbb{F}H|=p^{f|H|}=\prod_{i=1}^{n}|\mathbb{F}_i| = \prod_{i=1}^{n}p^{f_i}=p^{\sum_{i=1}^{n}f_i},$$
we have $f|H|=\sum_{i=1}^{n}f_i$.

Because $\mathbb{F}G = \mathbb{F}[G_p \oplus H] \cong (\mathbb{F}H)G_p \cong \bigoplus_{i=1}^n\mathbb{F}_iG_p$, we have the following decomposition
$$K_2(\mathbb{F}G) \cong \bigoplus_{i=1}^n K_2(\mathbb{F}_iG_p).$$
Since $|G^{p^i}|=|G_p^{p^i}||H|$, by Theorem~\ref{thm:mainp} and above decomposition we have
\begin{align*}
K_2(\mathbb{F}G) &= C_{p^e}^{(\sum_{j=1}^{n}f_j)(r_e-1)(|G_p^{p^{e-1}}|-1)}\bigoplus_{i=1}^{e-1}C_{p^i}^{(\sum_{j=1}^{n}f_j)\{(r_i-1)(|G_p^{p^{i-1}}|-|G_p^{p^i}|)-(r_{i+1}-1)(|G_p^{p^i}|-|G_p^{p^{i+1}}|)\}}\\
&= C_{p^e}^{f|H|(r_e-1)(|G_p^{p^{e-1}}|-1)}\bigoplus_{i=1}^{e-1}C_{p^i}^{f|H|\{(r_i-1)(|G_p^{p^{i-1}}|-|G_p^{p^i}|)-(r_{i+1}-1)(|G_p^{p^i}|-|G_p^{p^{i+1}}|)\}}\\
&=
C_{p^e}^{f(r_e-1)(|G^{p^{e-1}}|-|G^{p^{e}}|)}\bigoplus_{i=1}^{e-1}C_{p^i}^{f(r_i-1)(|G^{p^{i-1}}|-|G^{p^i}|)-(r_{i+1}-1)(|G^{p^i}|-|G^{p^{i+1}}|)}.
\end{align*}
\hspace{\fill}$\square$\vspace{0.3cm}


\section*{References}

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